NCERT Solutions: Class 12 Maths Chapter 10 - Vector Algebra

Choose a scale, draw north, then draw the vector from the initial point making $30^\circ$ towards east from north with scaled length $40$ km.

Only item (ii) states both magnitude and direction. The remaining measures, as written, give magnitude only.

Time period, distance and work done have magnitude only; force and velocity also have direction.

$\vec a$ and $-\vec a$ are parallel in opposite directions. Collinearity does not force equal magnitude, equal magnitude does not force collinearity, and same magnitude plus collinearity can still have opposite directions.

Use $|x\hat{i}+y\hat{j}+z\hat{k}|=\sqrt{x^2+y^2+z^2}$. This gives $\sqrt3$, $\sqrt{4+49+9}=\sqrt{62}$ and $\sqrt{1}=1$.

Both have magnitude $\sqrt2$, but the vectors are different.

They are positive scalar multiples of one another, so they have the same direction.

Equal vectors have equal corresponding components.

The vector is $(-5-2)\hat{i}+(7-1)\hat{j}=-7\hat{i}+6\hat{j}$.

Adding components gives $(1-2+1)\hat{i}+(-2+4-6)\hat{j}+(1+5-7)\hat{k}=-4\hat{j}-\hat{k}$.

The magnitude is $\sqrt{1+1+4}=\sqrt6$, so divide the vector by $\sqrt6$.

$\overrightarrow{PQ}=3\hat{i}+3\hat{j}+3\hat{k}$ and its magnitude is $3\sqrt3$.

$\vec a+\vec b=\hat{i}+\hat{k}$, whose magnitude is $\sqrt2$.

The magnitude of $5\hat{i}-\hat{j}+2\hat{k}$ is $\sqrt{30}$. Multiply its unit vector by $8$.

The second vector is $-2(2\hat{i}-3\hat{j}+4\hat{k})$, so one is a scalar multiple of the other.

The magnitude is $\sqrt{14}$; divide each component by this magnitude.

$\overrightarrow{AB}=-2\hat{i}-4\hat{j}+4\hat{k}$ and $|\overrightarrow{AB}|=6$. Divide each component by $6$.

Its direction cosines are all $1/\sqrt3$, so it makes equal angles with the three coordinate axes.

With $\vec p=\hat{i}+2\hat{j}-\hat{k}$ and $\vec q=-\hat{i}+\hat{j}+\hat{k}$, internal division gives $(2\vec q+\vec p)/3$, and external division gives $(2\vec q-\vec p)/(2-1)$.

The midpoint is $((2+4)/2,(3+1)/2,(4-2)/2)=(3,2,1)$.

$\overrightarrow{AB}=-\hat{i}+3\hat{j}+5\hat{k}$ and $\overrightarrow{AC}=-2\hat{i}+\hat{j}-\hat{k}$. Their dot product is $2+3-5=0$, so the angle at $A$ is $90^\circ$.

1. i. $\vec b=\lambda\vec a$, for some scalar $\lambda$
2. ii. $\vec a=\pm\vec b$
3. iii. the respective components of $\vec a$ and $\vec b$ are not proportional
4. iv. both the vectors $\vec a$ and $\vec b$ have same direction, but different magnitudes.

Collinear vectors satisfy $\vec b=\lambda\vec a$ for some scalar $\lambda$, so their components are proportional. They need not be equal/opposite or have the same direction.

$\cos\theta=\dfrac{\sqrt6}{2\sqrt3}=\dfrac1{\sqrt2}$, hence $\theta=\dfrac\pi4$.

The dot product is $10$ and both magnitudes are $\sqrt{14}$, so $\cos\theta=10/14=5/7$.

The dot product is $1-1=0$, so the scalar projection is $0$.

The dot product is $7-3+56=60$, and the magnitude of the second vector is $\sqrt{114}$.

The squared magnitude of each numerator vector is $49$, so each divided vector has magnitude $1$. The pairwise dot products are $6-18+12=0$, $12+6-18=0$ and $18-12-6=0$.

The first condition gives $|\vec a|^2-|\vec b|^2=8$. Put $|\vec b|=x$ and $|\vec a|=8x$ to get $63x^2=8$.

Distribute the dot product: $6\vec a\cdot\vec a+21\vec a\cdot\vec b-10\vec b\cdot\vec a-35\vec b\cdot\vec b$.

If the common magnitude is $x$, then $\vec a\cdot\vec b=x^2\cos60^\circ=x^2/2=1/2$, so $x=1$.

Expanding gives $|\vec x|^2-|\vec a|^2=12$. Since $|\vec a|=1$, $|\vec x|^2=13$.

Set $(\vec a+\lambda\vec b)\cdot\vec c=0$. This gives $3(2-\lambda)+(2+2\lambda)=8-\lambda=0$, so $\lambda=8$.

Their dot product is $|\vec a|^2|\vec b|^2-|\vec b|^2|\vec a|^2=0$.

$\vec a\cdot\vec a=|\vec a|^2=0$ implies $\vec a=\vec0$. Then $\vec a\cdot\vec b=0$ for every $\vec b$.

Squaring $\vec a+\vec b+\vec c=\vec0$ gives $3+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0$.

Nonzero perpendicular vectors have zero dot product, so zero dot product does not force one vector to be zero.

$\overrightarrow{BA}=(2,2,3)$ and $\overrightarrow{BC}=(1,1,2)$. Thus $\cos B=10/(\sqrt{17}\sqrt6)=10/\sqrt{102}$.

$\overrightarrow{AB}=(1,4,-4)$ and $\overrightarrow{BC}=(1,4,-4)$, so the points lie on the same line.

Taking these as position vectors $P,Q,R$, $\overrightarrow{RP}=-\hat{i}+3\hat{j}+5\hat{k}$ and $\overrightarrow{RQ}=-2\hat{i}+\hat{j}-\hat{k}$ have dot product $2+3-5=0$.

1. i. $\lambda=1$
2. ii. $\lambda=-1$
3. iii. $a=|\lambda|$
4. iv. $a=1/|\lambda|$

$|\lambda\vec a|=|\lambda|a$. For unit magnitude, $|\lambda|a=1$.

$\vec a\times\vec b=19\hat{j}+19\hat{k}$, so its magnitude is $19\sqrt2$.

A perpendicular vector is $(\vec a+\vec b)\times(\vec a-\vec b)=16\hat{i}-16\hat{j}-8\hat{k}$, whose magnitude is $24$.

The first two direction cosines are $1/2$ and $1/\sqrt2$. Hence $1/4+1/2+\cos^2\theta=1$, giving $\cos\theta=1/2$ as $\theta$ is acute.

Expanding gives $\vec a\times\vec a+\vec a\times\vec b-\vec b\times\vec a-\vec b\times\vec b=2(\vec a\times\vec b)$.

Zero cross product means the two vectors are parallel. Thus $(1,\lambda,\mu)=\dfrac12(2,6,27)$.

The dot product condition gives perpendicularity, while the cross product condition gives parallelism. Nonzero vectors cannot be both, so one vector must be zero.

Substitute components into the determinant formula for cross product; each component of $\vec a\times(\vec b+\vec c)$ separates into the corresponding components of $\vec a\times\vec b$ and $\vec a\times\vec c$.

The cross product of nonzero parallel vectors is zero, so the converse is false.

$\overrightarrow{AB}=(1,2,3)$ and $\overrightarrow{AC}=(0,4,3)$. Their cross product is $-6\hat{i}-3\hat{j}+4\hat{k}$, with magnitude $\sqrt{61}$. Triangle area is half of this.

$\vec a\times\vec b=20\hat{i}+5\hat{j}-5\hat{k}$, whose magnitude is $\sqrt{400+25+25}=15\sqrt2$.

1. i. $\pi/6$
2. ii. $\pi/4$
3. iii. $\pi/3$
4. iv. $\pi/2$

$|\vec a\times\vec b|=3(\sqrt2/3)\sin\theta=\sqrt2\sin\theta$. Setting this to $1$ gives $\sin\theta=1/\sqrt2$, so the listed answer is $\pi/4$.

1. i. $\dfrac12$
2. ii. $1$
3. iii. $2$
4. iv. $4$

The adjacent side lengths are $2$ and $1$, so the rectangle area is $2$.

A unit vector at angle $\theta$ in the XY-plane is $\cos\theta\hat{i}+\sin\theta\hat{j}$. Put $\theta=30^\circ$.

$\overrightarrow{PQ}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$.

Taking east and north as positive axes, the resultant displacement is $(-4,0)+(3\sin30^\circ,3\cos30^\circ)=(-5/2,3\sqrt3/2)$. Its magnitude is $\sqrt{13}$.

By triangle inequality $|\vec b+\vec c|\le |\vec b|+|\vec c|$. Equality holds only in the same-direction case; for example $|\hat{i}+\hat{j}|=\sqrt2\ne2$.

The magnitude is $|x|\sqrt3$. Setting it equal to $1$ gives $|x|=1/\sqrt3$.

The resultant is $3\hat{i}+\hat{j}$, with magnitude $\sqrt{10}$. Multiply its unit vector by $5$.

$2\vec a-\vec b+3\vec c=3\hat{i}-3\hat{j}+2\hat{k}$, whose magnitude is $\sqrt{22}$.

$\overrightarrow{AB}=(4,2,6)$ and $\overrightarrow{AC}=(10,5,15)=\frac52\overrightarrow{AB}$. Hence $AB:BC=2:3$.

With $\vec p=2\vec a+\vec b$ and $\vec q=\vec a-3\vec b$, external division gives $\vec r=(\vec q-2\vec p)/(1-2)=2\vec p-\vec q=3\vec a+5\vec b$. Also $(\vec r+\vec q)/2=\vec p$.

The diagonal vector is $3\hat{i}-6\hat{j}+2\hat{k}$ with magnitude $7$. The area is $|(2,-4,5)\times(1,-2,-3)|=|(22,11,0)|=11\sqrt5$.

Let the equal direction cosines be $l,l,l$. Then $3l^2=1$, giving $l=\pm1/\sqrt3$.

$\vec a\times\vec b=32\hat{i}-\hat{j}-14\hat{k}$. Put $\vec d=t(32\hat{i}-\hat{j}-14\hat{k})$. Then $\vec c\cdot\vec d=9t=15$, so $t=5/3$.

The sum vector is $(\lambda+2)\hat{i}+6\hat{j}-2\hat{k}$. The condition gives $(\lambda+6)/\sqrt{(\lambda+2)^2+40}=1$, so $(\lambda+6)^2=(\lambda+2)^2+40$ and $\lambda=1$.

Expanding gives $|\vec a|^2+2\vec a\cdot\vec b+|\vec b|^2$. Equality with $|\vec a|^2+|\vec b|^2$ is equivalent to $\vec a\cdot\vec b=0$.

1. i. $0\lt\theta\lt\dfrac\pi2$
2. ii. $0\le\theta\le\dfrac\pi2$
3. iii. $0\lt\theta\lt\pi$
4. iv. $0\le\theta\le\pi$

$\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$, so it is nonnegative when $\cos\theta\ge0$.

1. i. $\theta=\dfrac\pi4$
2. ii. $\theta=\dfrac\pi3$
3. iii. $\theta=\dfrac\pi2$
4. iv. $\theta=\dfrac{2\pi}{3}$

$|\vec a+\vec b|^2=2+2\cos\theta$. Setting this equal to $1$ gives $\cos\theta=-1/2$, so $\theta=2\pi/3$.

1. i. $0$
2. ii. $-1$
3. iii. $1$
4. iv. $3$

Since $\hat{j}\times\hat{k}=\hat{i}$, $\hat{i}\times\hat{k}=-\hat{j}$ and $\hat{i}\times\hat{j}=\hat{k}$, the value is $1-1+1=1$.

1. i. $0$
2. ii. $\dfrac\pi4$
3. iii. $\dfrac\pi2$
4. iv. $\pi$

The equality gives $|\cos\theta|=\sin\theta$. Among the choices, this holds at $\theta=\pi/4$.